Quiz: Electrical Engineering 12 Sep 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. D.C. shunt motors are used in those applications where ……… is required.
(a) high starting torque
(b) practically constant speed
(c) high no-load speed
(d) variable speed

Q2. Cumulatively compounded motors are used where we require ………
(a) variable speed
(b) poor speed regulation
(c) sudden heavy loads for short duration
(d) none of the above

Q3. For 20% increase in current, the motor that will give the greatest increase in torque
is ……. motor.
(a) shunt
(b) series
(c) cumulatively compounded
(d) differentially compounded

Q4. A 440 V shunt motor has an armature resistance of 0.8 Ω and a field resistance of 200
Ω. Find the back e.m.f. when giving an output of 7.46 kW at 85% efficiency.
(a) 222.4 V
(b) 425.8 V
(c) 312.6 V
(d) 392.7 V

Q5. A parallel resonance……….
(a) circuit impedance is minimum
(b) power factor is zero
(c) line current is maximum
(d) power factor is unity

Q6. Dynamic impedance of a parallel tuned circuit is ……….
(a) L/CR
(b) RL/C
(c) L/C
(d) R/L

Q7. When the supply frequency is more than the resonant frequency in a parallel a.c.
circuit, then circuit is…….
(a) Resistance
(b) capacitive
(c) inductive
(d) none of the above

Q8. A dc series motor drawing an armature current of I_a is operating under saturated magnetic conditions, torque developed in the motor is proportional to
(a) 1/Ia
(b) 1/Ia^2
(c) Ia^2
(d) Ia

Q9. The insulation resistance of a cable of length 10 Km is 1 MΩ. Its resistance for 50 Km length will be
(a) 1 MΩ
(b) 0.5 MΩ
(c) 0.2 MΩ
(d) None of these

Q10. For a circuit given that I = 2 ± 5 % A, R = 100 ± 0.2 % Ω the limiting error in the power dissipation I2 R in the resistor R is
(a) 1.2 %
(b) 5.2 %
(c) 10.2 %
(d) 25.2 %

SOLUTIONS
S1. Ans.(b)
Sol. Practically constant speed

S2. Ans.(c)
Sol. Cumulatively compounded motors are used where we require sudden heavy loads
for short duration

S3. Ans.(b)
Sol. Series motor because
T∝Ia^2 (Before saturation)

S4. Ans.(b)
Sol. Motor input power = (7.46×1000)/0.85 w
Motor input current = 7460/(0.85×440) = 19.95 A
Ish= 440/200 = 2.2 A
Ia= 19.95 – 2.2 = 17.75 A
Eb= V – Ia Ra= 440 – 17.75 × 0.8 = 425.8

S5. Ans.(d)
Sol. Power factor unity

S6. Ans.(a)
Sol. Dynamic impedance of parallel tuned circuit =L/RC

S7. Ans.(b)
Sol. Capacitive

S8. Ans.(d)
Sol. We know Torque Tα ΦIa, As the motor is operating under saturation condition so torque will be independent of flux i.e. T α Ia.

S9. Ans.(c)
Sol. Insulation resistance of the cable
R α 1/Length
R2/R1 = L1/L2
Hence, insulation resistance for 50 km length
R2 = (1 x 10)/50
= 0.2 MΩ

S10. Ans.(c)
Sol. Power dissipation P=I2R

Leave a comment

Your email address will not be published. Required fields are marked *